Question
Question: A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocit...
A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity w. Two objects each of mass m are attached gently to the opposite ends of a diameter of a ring. The ring now rotates with angular velocity given by
(A) (2mM+2m) w
(B) M+2m2Mw
(C) (MM+2m)w
(D) M+2mMw
Solution
Hint : We can use the law of conservation of angular momentum. According to this law initial angular momentum is equal to the final angular momentum of ring Li=Lf
Since, angular momentum is defined as a product of moment of inertia and angular velocity. So, Find the moment inertia of the ring before attaching masses and the moment of inertia of the ring after attaching masses.
Complete step by step answer
According to law of conservation of angular momentum
Li = Lf
Li → initial angular momentum of ring
Lf → final angular momentum of ring after attaching two objects of mass m, each.
Li = Iiw
Here, Iiis the moment of inertia of a ring without any object attached and w is angular velocity of object.
Ii = Mr2 here M is mass of ring and r is the radius of ring
∴Li = Mr2w ---(1)
Now, Lf=Ifw′
Here Ifis the moment of inertia of ring when two objects of mass m are attached, and angular velocity is denoted by w’
If=(M+2m)r2
∴Lf=(M+2m)r2w ------(2)
Since there is no external torque, we use conservation of angular momentum.
Equating the equation (1) and (2),
Li=Lf
Mr2w=(M+2m)r2w′
(M+2m)r2Mr2w=w′
w′=M+2mMw
This is the angular velocity with which ring attached with two objects of mass m rotates.
Note
Factors on which moment of inertia depends:
-Density of material
-Shape and size of body
-Axis of rotation
Law of conservation of angular momentum applied to
-Electric generator
-Aircraft engine etc.